it means EVERY element of R is of the form x + y, for some x in I, and y in J. in particular, 1 = x + y, for some x in I, and y in J (note: what you are trying to prove is NOT TRUE for rings without unity).

now it is clear that IJ is always contained in I∩J (see other thread for proof). what would it mean to say I∩J is contained in IJ?

well, it turns out we can do "algebra with ideals" (which simplifies having to consider just elements, as the expressions can get long and ugly).

the key result we will need is this:

(I+J)(I∩J) = I(I∩J) + J(I∩J)

suppose r is in (I+J)(I∩J).

then r = Σk (akbk), where ak is in I+J, for each k, and bk is in I∩J for each k.

thus we can write ak = xk + yk, where for each k, xk is in I, and yk is in J.

so r = Σk (akbk) = Σk ((xk+yk)bk) = Σk (xkbk) + Σk (ykbk),

which is clearly in I(I∩J) + J(I∩J).

on the other hand, suppose that r is instead in I(I∩J) + J(I∩J).

then r = Σk xkak + Σm ymbm, for xk in I, ym in J, and ak,bm in I∩J for each value of k and m.

suppose we let n = max(k,m), and add 0 terms for whichever sum comes up short. we can then re-write out sum as:

r = Σn (xnan + ynbn).

note that the term xjaj + yjbj is in (I+J)(I∩J), since xjaj = (xj+0)(aj), and yjbj = (0+yj)(bj),

and that this is true for EVERY j = 1,2...,n. since ideals are closed under addition, this means that r is in (I+J)(I∩J), so we have equality of (I+J)(I∩J) and I(I∩J) + J(I∩J).

ok, back to the main proof now:

note that for any ideal A, RA = A (clearly A is contained in RA, since for any a in A, a = 1a is in RA. on the other hand, an element of RA is r1a1+...+rnan, and each term rjaj is in A, since A is an ideal, so their sum is likewise in A...note how we needed a unity for R to assert this).